银行排队办业务问题
使用了 队列和结构体
算是比较麻烦的题了吧(可能我比较菜)

Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are:

  • The space inside the yellow line in front of each window is enough to contain a line with M customers. Hence when all the N lines are full, all the customers after (and including) the (NM+1)st one will have to wait in a line behind the yellow line.
  • Each customer will choose the shortest line to wait in when crossing the yellow line. If there are two or more lines with the same length, the customer will always choose the window with the smallest number.
  • Customeri will take Ti minutes to have his/her transaction processed.
  • The first N customers are assumed to be served at 8:00am.

Now given the processing time of each customer, you are supposed to tell the exact time at which a customer has his/her business done.

For example, suppose that a bank has 2 windows and each window may have 2 customers waiting inside the yellow line. There are 5 customers waiting with transactions taking 1, 2, 6, 4 and 3 minutes, respectively. At 08:00 in the morning, customer1 is served at window1 while customer2 is served at window2. customer3 will wait in front of window1 and customer4 will wait in front of window2. customer5 will wait behind the yellow line.

At 08:01, customer1 is done and customer5 enters the line in front of window1 since that line seems shorter now. Customer2 will leave at 08:02, customer4 at 08:06, customer3 at 08:07, and finally customer5 at 08:10.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers: N (≤20, number of windows), M (≤10, the maximum capacity of each line inside the yellow line), K (≤1000, number of customers), and Q (≤1000, number of customer queries).

The next line contains K positive integers, which are the processing time of the K customers.

The last line contains Q positive integers, which represent the customers who are asking about the time they can have their transactions done. The customers are numbered from 1 to K.

Output Specification:

For each of the Q customers, print in one line the time at which his/her transaction is finished, in the format HH:MM where HH is in [08, 17] and MM is in [00, 59]. Note that since the bank is closed everyday after 17:00, for those customers who cannot be served before 17:00, you must output Sorry instead.

Sample Input:

2 2 7 5
1 2 6 4 3 534 2
3 4 5 6 7

Sample Output:

08:07
08:06
08:10
17:00
Sorry

实现思想

银行排队问题

有N个窗口,每个窗口可排M个人

第N*M+1个及之后的要在黄线后等候,哪个窗口有人走并且人最少,去哪个窗口排队

四个窗口成为四个结构体对象,并存放在window

poptime为每个窗口第一个人离开的时间——用来判断应该去那个窗口排队

endtime为每个窗口最后一个人结束的时间——用来判断后面的人是否在17:00后输出Sorry

先将N*M个人的时间记录,然后再输入后面等待排队的人,并判断是Sorry是否为true

代码实现

#include <cstdio>
#include <queue>
#include <vector>

using namespace std;

struct node {
    int poptime, endtime;
    queue<int> q;
};

int main() {
    int n, m, k, q, index = 1;
    scanf("%d%d%d%d", &n, &m, &k, &q);
    vector<int> time(k + 1), result(k + 1);
    for(int i = 1; i <= k; i++) {
        scanf("%d", &time[i]);
    }
    vector<node> window(n + 1);
    vector<bool> sorry(k + 1, false);
    for(int i = 1; i <= m; i++) {
        for(int j = 1; j <= n; j++) {
            if(index <= k) {
                window[j].q.push(time[index]);
                if(window[j].endtime >= 540){
                    sorry[index] = true;
                }
                window[j].endtime += time[index];
                if(i == 1)
                    window[j].poptime = window[j].endtime;
                result[index] = window[j].endtime;
                index++;
            }
        }
    }

    while(index <= k) {
        int tempmin = window[1].poptime, tempwindow =1;
        for(int i = 2; i <= n; i++) {
            if(window[i].poptime < tempmin) {
                tempwindow = i;
                tempmin = window[i].poptime;
            }
        }
        window[tempwindow].q.pop();
        window[tempwindow].q.push(time[index]);
        window[tempwindow].poptime += window[tempwindow].q.front();
        if(window[tempwindow].endtime >= 540)
            sorry[index] = true;
        window[tempwindow].endtime += time[index];
        result[index] = window[tempwindow].endtime;
        index++;
    }

    for(int i = 1; i <= q; i++) {
        int query, minute;
        scanf("%d", &query);
        minute = result[query];
        if(sorry[query] == true) {
            printf("Sorry\n");
        } else {
            printf("%02d:%02d\n", (minute + 480) / 60, (minute + 480) % 60);
        }
    }

    return 0;
}